0195136047.pdf

364 SEMICONDUCTOR DEVICES

Current–

controlled

current

source

(fixed)

iB > 0

iC = βiB

iE = iC + iB

= (β + 1)iB

vCE > Vγ

vBE = Vγ

Vγ

−

−

+

+ +

E

(a)

C

B

iB > 0

iC < βiB

vCE = Vsat

Vsat

vBE = Vγ

(fixed)

Vγ

−

−

+

+ +

+

E

(b)

C

B

iB = 0

iC = ICEO

vCE ≥ 0

vBE < Vγ −

−

+

+

E

(c)

C

B

Figure 7.3.8Large-signal models ofnpnBJT.(a)Linear circuit model in idealized active state.(b)Idealized

saturated state.(c)Idealized cutoff state.

EXAMPLE 7.3.1

Consider the common-emitter BJT circuit shown in Figure E7.3.1(a). The static characteristics of

thenpnsilicon BJT are given in Figure E7.3.1(b) along with the load line. CalculateiBforvS= 1

V and 2 V. Then estimate the corresponding values ofvCEandiCfrom the load line, and compute

the voltage amplificationAv=vCE/vSand the current amplificationAi=iC/iB.

Solution

iB=0 forvS<VγandiB=(vS−Vγ)/RBforvS>V. With varying but positive base current,

vBEstays nearly constant at the junction threshold voltageVγ, which is 0.7 V for a silicon BJT

[see Figure 7.3.4(a)].

Then,IBQ 1 =

vS 1 − 0. 7

RB

=

1 − 0. 7

20,000

= 15 μA,forvS 1 = 0 .7V

Corresponding to 15-μA interpolated static curve and load line [see Fig. E7.3.1(b)], we get

vCE 1 = 9 .4 V, andiC 1 = 1 .3 mA;

forvS 2 =2V,IBQ 2 =

2 − 0. 7

20,000

=

1. 3

20,000

= 65 μA

+

−

C

(a)

E

+

+

−

+

− −

RB=Ω20 k B RC=Ω2 k

VCC=12 V

iC

iB

vS

vCE

vBE

Figure E7.3.1(a)Circuit.(b)Static

curves and load line.